eigenvalues-eigenvectorslinear algebramatrices
I will write an exam on Quantum Mechanics soon. I was wondering whether there is any smart and fast way to determine the eigenvalues/eigenvectors of a symmetric 3×3 matrix other than by calculating the characteristic polynomial?
So I am only intersted in fast techniques that one can use by hand to get those things.
I didn't notice the OP was asking for an easier way to unitary diagonalize a matrix. I'll leave this answer here, just because I find this algorithm interesting.
Regarding points 1) and 2), let me introduce an interesting algorithm to diagonalize a real symmetric matrix using only elementary row operations (taken from Schaum's Outline of Theory and Problems of Linear Algebra, by Lipschutz and Lipson).
And here is a worked example of the above mentioned algorithm:
The set of real numbers is a subset of the set of complex numbers, if we consider that real numbers are complex numbers with imaginary part equal to zero. Therefore, whatever holds for all complex numbers holds for real numbers.
As you observe, a polynomial might not have real roots. However, all the eigenvalues of a symmetric real matrix are real. By definition, they are the roots of the characteristic polynomial, so you can be sure that an example like the one you proposed will not arise.
Best Answer
There is no special trick to find the eigenvalues/vectors of your matrix $A$ (instead of you just "see them" or more special cases of $A$) but if you want to diagonalize your matrix you can make benefit of the symmetry of the matrix when finding an transformation matrix $S$ with $\Delta = S A S^{-1} $
If $A$ is symmetric you know, that eigenvectors to different eigenvalues are ortthogonal, so it's most time very easy to find even an orthonormal basis of eigenvalues. The benefit for your calculation then is, that you don't have to find the inverse of your transformation matrix $S$ because it is given by its transposed matrix $S^T$
Now you are able to calculate the diagonal matrix $\Delta$ simply by $\Delta = S^T A S$